Thread: Geometric Random Variable

Let W be the number of failures until the first success; that is, W = T-1. Show that W has mean (1-pi)/pi and variance (1-pi)/(pi^2).

I know for mean, E(T) = 1/pi and for variance Var(T) = (1-pi)/(pi^2).

I guess I'm looking for E(T(W)) = E(T(T-1)) and Var(T(W)) = Var(T(T-1)).

I've included how I've started so far in the attachment. Note I messed up on the last part in the step where I moved pi in front of summation. Should say (t-1). Sorry.

Can someone help me with this proof.

Originally Posted by DINOCALC09

Let W be the number of failures until the first success; that is, W = T-1. Show that W has mean (1-pi)/pi and variance (1-pi)/(pi^2).

I know for mean, E(T) = 1/pi and for variance Var(T) = (1-pi)/(pi^2).

I guess I'm looking for E(T(W)) = E(T(T-1)) and Var(T(W)) = Var(T(T-1)).

I've included how I've started so far in the attachment. Note I messed up on the last part in the step where I moved pi in front of summation. Should say (t-1). Sorry.

Can someone help me with this proof.

You will find the derivation of the mean and variance of a geometric random variable in most textbook on Mathematical Statistics. Here are some websites (found using Google in under 5 minutes):

http://www.win.tue.nl/~rnunez/2DI30/...tributions.pdf

Expectation and variance of the discrete geometric distribution. · Digital explorations

In my books it shows E[T] = 1/pi

In the attachments it shows E[T] = 1/(1-pi)

A little confused...

PS, is my setp correct thus far?

Originally Posted by DINOCALC09

In my books it shows E[T] = 1/pi

In the attachments it shows E[T] = 1/(1-pi)

A little confused...

PS, is my setp correct thus far?

There are two different definitions for the geometric distribution: see Geometric distribution - Wikipedia, the free encyclopedia

You have to decide which one is being used in your problem.

So...

The probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, ... } according to Wikipedia.

Now, how do I change the summation in the next step to t=2 as the lower bounds. This is the step I'm stuck on.

Originally Posted by DINOCALC09

So...

The probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, ... } according to Wikipedia.

Now, how do I change the summation in the next step to t=2 as the lower bounds. This is the step I'm stuck on.

Did you read the links I gave you? The derivation is given.

ok i get it now. will post my work once i'm done working it out on paper. thank you thank you thank you